Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation 3*x^2-8*x-3=0
-
Equation x^3=9
-
Equation z^4+z^2+1=0
- Equation z^3=-i
- Express {x} in terms of y where:
- -10*x+5*y=10
- -18*x+16*y=4
- 6*x+9*y=15
- -8*x-15*y=-12
- z^3
- Derivative of:
-
z^3
- Graphing y =:
- z^3
-
Identical expressions
- z^ three =-i
- z cubed equally minus i
- z to the power of three equally minus i
- z3=-i
- z³=-i
- z to the power of 3=-i
-
Similar expressions
- z^3=+i
z^3=-i equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$z^{3} = - i$$
Because equation degree is equal to = 3 - does not contain even numbers in the numerator, then
the equation has single real root.
Get the root 3-th degree of the equation sides:
We get:
$$\sqrt[3]{z^{3}} = \sqrt[3]{- i}$$
or
$$z = \sqrt[3]{- i}$$
Expand brackets in the right part
We get the answer: z = (-i)^(1/3)
All other 3 root(s) is the complex numbers.
do replacement:
$$w = z$$
then the equation will be the:
$$w^{3} = - i$$
Any complex number can presented so:
$$w = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = - i$$
where
$$r = 1$$
- the magnitude of the complex number
Substitute r:
$$e^{3 i p} = - i$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(3 p \right)} + \cos{\left(3 p \right)} = - i$$
so
$$\cos{\left(3 p \right)} = 0$$
and
$$\sin{\left(3 p \right)} = -1$$
then
$$p = \frac{2 \pi N}{3} - \frac{\pi}{6}$$
where N=0,1,2,3,...
Looping through the values of N and substituting p into the formula for w
Consequently, the solution will be for w:
$$w_{1} = i$$
$$w_{2} = - \frac{\sqrt{3}}{2} - \frac{i}{2}$$
$$w_{3} = \frac{\sqrt{3}}{2} - \frac{i}{2}$$
do backward replacement
$$w = z$$
$$z = w$$
The final answer:
$$z_{1} = i$$
$$z_{2} = - \frac{\sqrt{3}}{2} - \frac{i}{2}$$
$$z_{3} = \frac{\sqrt{3}}{2} - \frac{i}{2}$$
$$z^{3} = - i$$
Because equation degree is equal to = 3 - does not contain even numbers in the numerator, then
the equation has single real root.
Get the root 3-th degree of the equation sides:
We get:
$$\sqrt[3]{z^{3}} = \sqrt[3]{- i}$$
or
$$z = \sqrt[3]{- i}$$
Expand brackets in the right part
z = -i^1/3
We get the answer: z = (-i)^(1/3)
All other 3 root(s) is the complex numbers.
do replacement:
$$w = z$$
then the equation will be the:
$$w^{3} = - i$$
Any complex number can presented so:
$$w = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = - i$$
where
$$r = 1$$
- the magnitude of the complex number
Substitute r:
$$e^{3 i p} = - i$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(3 p \right)} + \cos{\left(3 p \right)} = - i$$
so
$$\cos{\left(3 p \right)} = 0$$
and
$$\sin{\left(3 p \right)} = -1$$
then
$$p = \frac{2 \pi N}{3} - \frac{\pi}{6}$$
where N=0,1,2,3,...
Looping through the values of N and substituting p into the formula for w
Consequently, the solution will be for w:
$$w_{1} = i$$
$$w_{2} = - \frac{\sqrt{3}}{2} - \frac{i}{2}$$
$$w_{3} = \frac{\sqrt{3}}{2} - \frac{i}{2}$$
do backward replacement
$$w = z$$
$$z = w$$
The final answer:
$$z_{1} = i$$
$$z_{2} = - \frac{\sqrt{3}}{2} - \frac{i}{2}$$
$$z_{3} = \frac{\sqrt{3}}{2} - \frac{i}{2}$$
Vieta's Theorem
it is reduced cubic equation
$$p z^{2} + q z + v + z^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = i$$
Vieta Formulas
$$z_{1} + z_{2} + z_{3} = - p$$
$$z_{1} z_{2} + z_{1} z_{3} + z_{2} z_{3} = q$$
$$z_{1} z_{2} z_{3} = v$$
$$z_{1} + z_{2} + z_{3} = 0$$
$$z_{1} z_{2} + z_{1} z_{3} + z_{2} z_{3} = 0$$
$$z_{1} z_{2} z_{3} = i$$
$$p z^{2} + q z + v + z^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = i$$
Vieta Formulas
$$z_{1} + z_{2} + z_{3} = - p$$
$$z_{1} z_{2} + z_{1} z_{3} + z_{2} z_{3} = q$$
$$z_{1} z_{2} z_{3} = v$$
$$z_{1} + z_{2} + z_{3} = 0$$
$$z_{1} z_{2} + z_{1} z_{3} + z_{2} z_{3} = 0$$
$$z_{1} z_{2} z_{3} = i$$
The graph
Rapid solution
[src]
z1 = I
$$z_{1} = i$$
___
I \/ 3
z2 = - - - -----
2 2
$$z_{2} = - \frac{\sqrt{3}}{2} - \frac{i}{2}$$
___
\/ 3 I
z3 = ----- - -
2 2
$$z_{3} = \frac{\sqrt{3}}{2} - \frac{i}{2}$$
z3 = sqrt(3)/2 - i/2
Sum and product of roots
[src]
sum
___ ___
I \/ 3 \/ 3 I
I + - - - ----- + ----- - -
2 2 2 2
$$\left(\frac{\sqrt{3}}{2} - \frac{i}{2}\right) + \left(\left(- \frac{\sqrt{3}}{2} - \frac{i}{2}\right) + i\right)$$
=
0
$$0$$
product
/ ___\ / ___ \ | I \/ 3 | |\/ 3 I| I*|- - - -----|*|----- - -| \ 2 2 / \ 2 2/
$$i \left(- \frac{\sqrt{3}}{2} - \frac{i}{2}\right) \left(\frac{\sqrt{3}}{2} - \frac{i}{2}\right)$$
=
-I
$$- i$$
-i
Numerical answer
[src]
z1 = 0.866025403784439 - 0.5*i
z2 = 1.0*i
z3 = -0.866025403784439 - 0.5*i
z3 = -0.866025403784439 - 0.5*i